以 O(1) 时间复杂度和 O(1) 额外空间在 C++ 中查找栈中的最大值
假设我们想创建一个堆栈,该堆栈可以存储堆栈中的最大元素。并且我们可以在 O(1) 时间内获取它。限制条件是,它应该使用 O(1) 额外空间。
我们可以创建一个自定义堆栈,它将存储最大值,当执行一个操作(例如 pop 或 peek)时,将返回最大值。对于 peek 操作,返回堆栈顶部和最大元素的最大值,对于 pop 操作,当顶部元素较大时,打印它并将最大值更新为 2*max – top_element。否则,返回顶部元素。对于 push 操作,将最大元素更新为 x(要插入的数据),2*x – max。
示例
#include <iostream> #include <stack> using namespace std; class CustomStack { stack<int> stk; int stack_max; public: void getMax() { if (stk.empty()) cout << "Stack is empty"<<endl; else cout << "Maximum Element in the stack is: "<< stack_max <<endl; } void peek() { if (stk.empty()) { cout << "Stack is empty "; return; } int top = stk.top(); // Top element. cout << "Top Most Element is: "<<endl; (top > stack_max) ? cout << stack_max : cout << top; } void pop() { if (stk.empty()) { cout << "Stack is empty"<<endl; return; } cout << "Top Most Element Removed: "; int top = stk.top(); stk.pop(); if (top > stack_max) { cout << stack_max <<endl; stack_max = 2 * stack_max - top; } else cout << top <<endl; } void push(int element) { if (stk.empty()) { stack_max = element; stk.push(element); cout << "Element Inserted: " << element <<endl; return; } if (element > stack_max) { stk.push(2 * element - stack_max); stack_max = element; } else stk.push(element); cout << "Element Inserted: " << element <<endl; } }; int main() { CustomStack stk; stk.push(4); stk.push(6); stk.getMax(); stk.push(8); stk.push(20); stk.getMax(); stk.pop(); stk.getMax(); stk.pop(); stk.peek(); }
输出
Element Inserted: 4 Element Inserted: 6 Maximum Element in the stack is: 6 Element Inserted: 8 Element Inserted: 20 Maximum Element in the stack is: 20 Top Most Element Removed: 20 Maximum Element in the stack is: 8 Top Most Element Removed: 8 Top Most Element is: 6
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